Series resistance of battery pack?

[Joe]

Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a breadboard I
have constructed, but I have no idea what the series resistance of a battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe

 

[CFoley1064]

Subject: Series resistance of battery pack?
From: "Joe" nuisancewildlife@nospamearthlink.net
Date: 2/25/2004 5:53 PM Central Standard Time
Message-id: <jAa%b.21612$hm4.9368@newsread3.news.atl.earthlink.net

Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a breadboard I
have constructed, but I have no idea what the series resistance of a battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe

Choose a resistor which will load the battery pack to something near
recommended maximum current. After charging the battery pack, measure no-load
voltage with a DMM (Vnl). Place the resistor across the battery pack, and
measure voltage across the resistor (Vfl). Calculate current across your load
resistor

(I = V/R).

Then calculate internal resistance by

R = (Vnl - Vfl) / I.

Good luck
Chris

 

[Costas Vlachos]

"Joe" <nuisancewildlife@nospamearthlink.net> wrote in message
news:jAa%b.21612$hm4.9368@newsread3.news.atl.earthlink.net...

Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a breadboard
I
have constructed, but I have no idea what the series resistance of a
battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe

Measure the battery's voltage without any load. Call this value V1. Connect
an external load resistor across the battery and measure the battery's new
voltage and the current flowing through the load. Call these values V2 and
I. The battery's internal resistance is given by:

Ri = (V1 - V2) / I

Make sure you use a resistor that generates enough current to cause a
significant drop in the battery's voltage.

This assumes a battery model consisting of an ideal voltage source and a
resistor. For a more accurate measurement you'll need to plot a graph of
battery voltage vs. load current and evaluate the slope of the graph at the
operating point you need.

cheers,
Costas
_________________________________________________
Costas Vlachos Email: c-X-vlachos@hot-X-mail.com
SPAM-TRAPPED: Please remove "-X-" before replying

 

[Joe]

"CFoley1064" <cfoley1064@aol.com> wrote in message
news:20040225193609.22806.00000376@mb-m10.aol.com...

Subject: Series resistance of battery pack?
From: "Joe" nuisancewildlife@nospamearthlink.net
Date: 2/25/2004 5:53 PM Central Standard Time
Message-id: <jAa%b.21612$hm4.9368@newsread3.news.atl.earthlink.net

Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a breadboard
I
have constructed, but I have no idea what the series resistance of a
battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe

Choose a resistor which will load the battery pack to something near
recommended maximum current. After charging the battery pack, measure
no-load
voltage with a DMM (Vnl). Place the resistor across the battery pack, and
measure voltage across the resistor (Vfl). Calculate current across your
load
resistor

(I = V/R).

Then calculate internal resistance by

R = (Vnl - Vfl) / I.

Good luck
Chris

Chris,

Thank you for the reply. I am not sure what is meant by recommended maximum
current. I have a battery chart that shows the mAH for different types of
batteries at a certain current drain value, eg, for AA alkaline manganese
dioxide batteries, the chart shows 2000mAH at a typical drain of 50mA.
Is that what I use (50mA)?

Joe

 

[Joe]

"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote in message
news:c1jfor$fgl$1@titan.btinternet.com...

"Joe" <nuisancewildlife@nospamearthlink.net> wrote in message
news:jAa%b.21612$hm4.9368@newsread3.news.atl.earthlink.net...
Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a
breadboard
I
have constructed, but I have no idea what the series resistance of a
battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe


Measure the battery's voltage without any load. Call this value V1.
Connect
an external load resistor across the battery and measure the battery's new
voltage and the current flowing through the load. Call these values V2 and
I. The battery's internal resistance is given by:

Ri = (V1 - V2) / I

Make sure you use a resistor that generates enough current to cause a
significant drop in the battery's voltage.

This assumes a battery model consisting of an ideal voltage source and a
resistor. For a more accurate measurement you'll need to plot a graph of
battery voltage vs. load current and evaluate the slope of the graph at
the
operating point you need.

cheers,
Costas
snip

Thank you Costas, a significant drop in battery voltage, like maybe it will
drop the voltage by one half? Would that be enough?

Thanks,
Joe

 

[John Popelish]

Joe wrote:

Thank you for the reply. I am not sure what is meant by recommended maximum
current. I have a battery chart that shows the mAH for different types of
batteries at a certain current drain value, eg, for AA alkaline manganese
dioxide batteries, the chart shows 2000mAH at a typical drain of 50mA.
Is that what I use (50mA)?

That is a reasonable load, especially if it is close to what your
future load will be. You might measure the voltage of the pack before
during and after applying that load, and average the before and after
as the unloaded voltage. Subtract the loaded voltage from that
average and divide that difference by the load current (in amperes,
..05 in this case) to calculate the approximate battery resistance.
--
John Popelish

 

[Costas Vlachos]

"Joe" <nuisancewildlife@nospamearthlink.net> wrote in message
news:3ny%b.23628$hm4.20414@newsread3.news.atl.earthlink.net...

"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote in message
news:c1jfor$fgl$1@titan.btinternet.com...

Measure the battery's voltage without any load. Call this value V1.
Connect an external load resistor across the battery and measure the
battery's new voltage and the current flowing through the load. Call
these values V2 and I. The battery's internal resistance is given by:

Ri = (V1 - V2) / I

Make sure you use a resistor that generates enough current to cause a
significant drop in the battery's voltage.

This assumes a battery model consisting of an ideal voltage source and a
resistor. For a more accurate measurement you'll need to plot a graph of
battery voltage vs. load current and evaluate the slope of the graph at
the operating point you need.

cheers,
Costas
snip

Thank you Costas, a significant drop in battery voltage, like maybe it
will drop the voltage by one half? Would that be enough?

Thanks,
Joe

One half sounds a bit too much. How about setting the current to the nominal
current of your circuit? That would give you a more accurate Rin masurement
as the relationship may not be linear depending on the type of battery, etc.

Costas

 

[David Wood]

It might be interesting to compare your results with reference data
from the cell manufacturer. For example, the datasheet for the
ubiquitous Duracell Alkaline-Manganese AA shows 120 m-Ohm at 1kHz:

http://www.duracell.com/oem/Pdf/MN1500.pdf

Why would they reference a frequency rather than DC in their chart?

 

[Joe]

"John Popelish" <jpopelish@rica.net> wrote in message
news:403EB73E.E2B1A813@rica.net...

Joe wrote:

Thank you for the reply. I am not sure what is meant by recommended
maximum
current. I have a battery chart that shows the mAH for different types
of
batteries at a certain current drain value, eg, for AA alkaline
manganese
dioxide batteries, the chart shows 2000mAH at a typical drain of 50mA.
Is that what I use (50mA)?

That is a reasonable load, especially if it is close to what your
future load will be. You might measure the voltage of the pack before
during and after applying that load, and average the before and after
as the unloaded voltage. Subtract the loaded voltage from that
average and divide that difference by the load current (in amperes,
.05 in this case) to calculate the approximate battery resistance.
--
John Popelish

Thanks John, I measured a rechargeable 9V (actually 7.2Volt) NiMH battery
to start with. I came up with about 6ohms. Is that about what you would
expect? I measured it with 70mA (approx 100 ohm) load. It is not fully
charged for probly a month or so, just been sitting on my bench, but it has
not been used much either.

I am going to be using a 12volt lawnmower battery to power one of my
projects and that is on the charger right now, so I can measure it when it
is at full charge and keep track of it as it discharges.

Joe

 

[Joe]

"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote in message
news:c1nlbi$5rj$1@sparta.btinternet.com...

"Joe" <nuisancewildlife@nospamearthlink.net> wrote in message
news:3ny%b.23628$hm4.20414@newsread3.news.atl.earthlink.net...

"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote in message
news:c1jfor$fgl$1@titan.btinternet.com...

Measure the battery's voltage without any load. Call this value V1.
Connect an external load resistor across the battery and measure the
battery's new voltage and the current flowing through the load. Call
these values V2 and I. The battery's internal resistance is given by:

Ri = (V1 - V2) / I

Make sure you use a resistor that generates enough current to cause a
significant drop in the battery's voltage.

This assumes a battery model consisting of an ideal voltage source and
a
resistor. For a more accurate measurement you'll need to plot a graph
of
battery voltage vs. load current and evaluate the slope of the graph
at
the operating point you need.

cheers,
Costas
snip

Thank you Costas, a significant drop in battery voltage, like maybe it
will drop the voltage by one half? Would that be enough?

Thanks,
Joe


One half sounds a bit too much. How about setting the current to the
nominal
current of your circuit? That would give you a more accurate Rin
masurement
as the relationship may not be linear depending on the type of battery,
etc.

Costas

Hi Costas,

I did that and came up with about 6ohms for a rechargeable NiMH (7.2V). I am
using a different battery for my application so I will measure that one
when it is fully charged and possibly plot the values for a learning
experience. Thanks for the tip

Joe

 

[Joe]

"David Wood" <davidwood911@yahoo.com> wrote in message
news:90c30deb.0402271649.203bd1f9@posting.google.com...

It might be interesting to compare your results with reference data
from the cell manufacturer. For example, the datasheet for the
ubiquitous Duracell Alkaline-Manganese AA shows 120 m-Ohm at 1kHz:

http://www.duracell.com/oem/Pdf/MN1500.pdf

Why would they reference a frequency rather than DC in their chart?

Hi David,

I was wondering that myself. I went to the site, thanks for the link. The
graphs they show are very instructive, but I don't know what's up with the
impedance at 1Khz, maybe someone else on this forum can explain it...

Joe

 

[John Popelish]

"Joe" <nuisancewildlife@nospamearthlink.net> wrote in message news:<xIR%b.25019$hm4.7789@newsread3.news.atl.earthlink.net>...

Thanks John, I measured a rechargeable 9V (actually 7.2Volt) NiMH battery
to start with. I came up with about 6ohms.

That sounds quite reasonable to me. That would correspond to a >1 amp
short circuit current.

Is that about what you would
expect? I measured it with 70mA (approx 100 ohm) load. It is not fully
charged for probly a month or so, just been sitting on my bench, but it has
not been used much either.

I am going to be using a 12volt lawnmower battery to power one of my
projects and that is on the charger right now, so I can measure it when it
is at full charge and keep track of it as it discharges.

That one should measure well under and ohm.

--
John Popelish